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 October 14th, 2015, 09:57 PM #1 Newbie   Joined: Oct 2015 From: England Posts: 2 Thanks: 0 Projection Transformation on x Axis Parallel to y=2x Hello, Let $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the projection transformation on x axis in parallel to $y=2x$. Find the representing matrix of $T^*$ for the standard basis $B=\left(\left \{1,0 \right \},\left\{0,1\right\}\right)$. I have a solution which I don't understand. A basis for $y=2x$ could be $\left \{(1,2) \right\}$. A basis for x axis could be $\left \{(1,0)\right\}$. Therefore A basis for the whole vector would be $$C=\left\{(1,0)\right\},\left\{(1,2)\right\}$$. I understand why this basis is correct. However: $e_1=(1,0)=1v_1+0v_2 \Rightarrow [e_1]_B=\left (\begin {matrix} 1\\ 0 \end {matrix}\right)$ - this is quite clear. $e_2=(0,1)=-\frac{1}{2}v_1+\frac{1}{2}v_2\Rightarrow [e_2]_B=\left (\begin {matrix} -\frac{1}{2} \\ \frac{1}{2} \end {matrix}\right)$ - this is the part I don't understand. Why use $-\frac{1}{2}$ to multiply $v_1$? What am I missing here? Next, since the proejction transformation is $T(x,y)=(x,0)$ it is easy enough to continue. But there's that part I don't understand. Why is the scalar before the vector $v_1$has to be$-\frac{1} {2}$ Thanks, Alan October 15th, 2015, 09:20 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Given unit vector u. Projection of any vector x onto u in matrix form is given by Tx where $\displaystyle T_{ij} =u_{i}u_{j}$ Change of basis: e1'=P11e1+P21e2 e2'=P12e1+P22e2 Relative to new basis: $\displaystyle T'=P^{-1}TP$ Reference: https://www.khanacademy.org/math/lin...ix-vector-prod Linear Transformations. EDIT: I think I see now what you are asking: Find the projection of x onto x axis in direction of line y=2x. Have to think about how to put that in matrix form. Last edited by zylo; October 15th, 2015 at 09:28 AM. October 15th, 2015, 10:08 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Projection of x in direction u on x axis. ae1+bu=x and solve for a a=x1-x2u1/u2 In matrix form: Tx= $\displaystyle \begin{bmatrix} 1 &-u1/u2 \\ 0& 0 \end{bmatrix}\begin{bmatrix} x1\\ x2 \end{bmatrix}=\begin{bmatrix} x1-x2u1/u2\\ 0 \end{bmatrix}$ Last edited by zylo; October 15th, 2015 at 10:20 AM. October 19th, 2015, 09:47 AM #4 Newbie   Joined: Oct 2015 From: England Posts: 2 Thanks: 0 Thank you both! October 20th, 2015, 08:33 AM   #5
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Quote:
 Originally Posted by Nerazzurri10 Thank you!
You're welcome. October 31st, 2015, 07:17 AM #6 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Systematic Solution to OP based on first principles. T:Rn->Rm is a linear transformation if T(au+bv)=aTu+bTv Matrix equivalent of T:R2->R2 T(x1e1+x2e2)=x1Te1+x2Te2=x1(T11e1+T21e2)+x2(T12e1+ T22e2) Tij are the matrix components of T Tx=[T]x Example: Project x onto line w in direction v u: unit vector in direction of x v: unit vector in direction of projection w: unit vector along line which x is to be projected onto, au+bv=cw (draw a sketch, all vectors from origin) Te1=cw when au=e1 e1+bv=cw 1+bv1=cw1 bv2=cw2, and solve: c=v2/(w1v2-w2v1) Te2=c'w when a'u=e2 e2+b'v=c'w b'v1=c'e1 1+b'v2=c'w2, and solve c'=-v1/(w1v2-w2v1) Te1=cw=c(w1e1+w2e2)=T11e1+T21e2 Te2=c'w=c'(w1e1+w2e2)=T12e1+T22e2 T11=cw1, T21=cw2, T12=c'w1, T22=c'w2 EDIT: That's the general case. For OP line to be projected onto is x axis so w1=1, w2=0, and also v1=1/sqrt(5), v2=2/sqrt(5) Last edited by zylo; October 31st, 2015 at 07:30 AM. Tags axis, parallel, projection, transformation, y2x Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathmaniac Physics 15 March 18th, 2015 07:14 AM brhum Pre-Calculus 4 November 27th, 2014 04:16 AM Drasik Geometry 4 October 11th, 2014 07:08 AM foensje Calculus 2 March 22nd, 2014 02:39 PM Anamitra Palit Physics 3 December 6th, 2012 01:20 PM

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